Problem: $\overline{AC}$ is $12$ units long $\overline{BC}$ is $5$ units long $\overline{AB}$ is $13$ units long What is $\csc(\angle BAC)?$ $A$ $C$ $B$ $12$ $5$ $13$
Solution: $\csc(\angle BAC) = \dfrac{1}{\sin(\angle BAC)}$ How can we find $\sin(\angle BAC)$ SOH CAH TOA in = pposite over ypotenuse Opposite $= \overline{BC} = 5$ Hypotenuse $= \overline{AB} = 13$ $\sin(\angle BAC) = \dfrac{5}{13}$ $\csc(\angle BAC) = \dfrac{1}{\sin(\angle BAC)} = \dfrac{13}{5}$